Answer
The solutions of the equation are $x=\frac{5\pi }{4}+2n\pi $ or $x=\frac{7\pi }{4}+2n\pi $, where n is any integer.
Work Step by Step
The x in $sinx=-\frac{\sqrt{2}}{2}$. Because $sin\frac{\pi }{4}=\frac{\sqrt{2}}{2}$, and the solutions of $sinx=-\frac{\sqrt{2}}{2}$ in [0,2 $\pi $ ) are:
$\begin{align}
& x=\pi +\frac{\pi }{4} \\
& =\frac{4\pi }{4}+\frac{\pi }{4} \\
& =\frac{5\pi }{4}
\end{align}$
$\begin{align}
& x=2\pi -\frac{\pi }{4} \\
& =\frac{8\pi }{4}-\frac{\pi }{4} \\
& =\frac{7\pi }{4}
\end{align}$
The period of the sine function is $2\pi $, and the solutions are given by:
$x=\frac{5\pi }{4}+2n\pi $ or $x=\frac{7\pi }{4}+2n\pi $, where n is any integer.
